SOLUTION: Approximately 15% of the calls to an airline reservation phone line result in a reservation being made. What is the probability that at least one call results in a reservation bein
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Question 349766: Approximately 15% of the calls to an airline reservation phone line result in a reservation being made. What is the probability that at least one call results in a reservation being made? (Give the answer to 3 decimals places.)
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
Can't answer this specifically because you didn't share the number of trials.
The probability of 
successes in 
trials where 
is the probability of success on any given trial is given by:
\ =\ \left(n\cr k\right\)p^k\left(1\,-\,p\right)^{n\,-\,k})
Where )
is the number of combinations of things taken at a time and is calculated by !})
You have 
and you have is at least 1. But you didn't give .
The probability of at least 1 is the probability of exactly 1 plus the probability of exactly 2 plus exactly 3...and so on up to the probability of exactly out of . Such a calculation would look like this:
\ =\ \sum_{i=1}^n\left(n\cr i\right\)\left(0.15\right)^i\left(0.85\right)^{n\,-\,i})
The calculation of which, depending on the size of , could range anywhere from mildly uncomfortable to excruciating. Ah, but take heart, Grasshopper! There is a better way.
Note that if "at least 1" is one possibility, then the only other possiblity is "none at all". Hence the probability of "at least one" plus the probability of "none at all" must equal 1. Therefore, if you calculate the probability of none at all and subtract that from 1, you get your desired probability.
\ =\ 1\ -\ P_n(0,0.15)\ =\ 1\ -\ \left(n\cr 0\right\)\left(0.15\right)^0\left(0.85\right)^n)
.
But since \ =\ 1)
and 
, this all reduces to:
\ =\ 1\ -\ P_n(0,0.15)\ =\ 1\ -\ \left(0.85\right)^n)
.
All you need to do is determine and then do the arithmetic.
John
My calculator said it, I believe it, that settles it
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